Question

taylor expansion question! need help! thank you!

Answer 1

@amistre64 can you help me this problem??

Answer 2

given that
\[\int_{0}^{x} sin(t)~dt=1-cos(x)\]
rearrange
\[cos(x)=1-\int_{0}^{x} sin(t)~dt\]
you are given the polynomial expansion of sin, so replace sin with it
\[cos(x)=1-\int_{0}^{x} f(t)~dt\]

Answer 3

here is my step, but my instructor said it's wrong, i don't know what's going on

Answer 4

for starters, F(x)-F(0) is not F(t)

Answer 5

\[\int_{0}^x(\frac{t}{1!}-\frac{t^3}{3!}+\frac{t^5}{5!}-...+\frac{(-1)^kt^{2k+1}}{(2k+1)!}+...)dt~:~k=0,1,2,3,...\]
\[\left.\frac{t^2}{2*1!}-\frac{t^4}{4*3!}+\frac{t^6}{6*5!}-...+\frac{(-1)^kt^{2k+2}}{(2x+2)*(2k+1)!}+...\right|_{0}^{x}\]

Answer 6

i change it like this, it that correct now?

Answer 7

when t=0; the -F(0) = 0, so lets focus on the F(x); and adjust out the typos :)

\[\frac{x^2}{2!}-\frac{x^4}{4!}+\frac{x^6}{6!}-...+\frac{(-1)^kx^{2k+2}}{(2k+2)*(2k+1)!}+...\]
2k+2 = 2(k+2), which is an even number; lets say 2n

\[\frac{x^2}{2!}-\frac{x^4}{4!}+\frac{x^6}{6!}-...+\frac{(-1)^kx^{2n}}{(2k+2)*(2k+1)!}+...\]
and you got the 1 - in there correctly

can we show that: (2k+2)*(2k+1)! = (2n)! ?

Answer 8

2(k+1) = 2n in that middle part ....

Answer 9

hmmm, i wonder if adjustments "by pattern" would be acceptable to your teacher

Answer 10

what do you mean by" (2k+2)*(2k+1)! = (2n)! "

Answer 11

\[\int t^{2k+1}dt=\frac{t^{(2k+1)+1}}{(2k+1)+1}\]right?

Answer 12

yes

Answer 13

and since all our denominators are turning out to be even factorials ... i was wondering if there was a simple way to show that; at least\[(2k+2)*(2k+1)!=(2k+2)!~or~(2(k+1))!\]

Answer 14

by definition of factorials:\[(n+1)*n!=(n+1)!\]
by definition of factorials:\[((2k+1)+1)*(2k+1)!=((2k+1)+1)!\]
should work, unless they want you to reinvent the wheel

Answer 15

so the last term should be 2n! or ?

Answer 16

it should be an even term, yes; and 2k+2 factors to 2(k+1) which is the definition of an even number

Answer 17

if we let: n=(k+1), then (n-1)=k, giving us
\[cos(x) = 1-(\sum_{n=1}^{\infty}\frac{(-1)^{n-1}x^{2n}}{(2n)!})\]